Variation of Parameters Calculator

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Example

Created on 2024-06-20Asked by Sofia Smith (Solvelet student)

Solve the differential equation

y'' - y = e^x

using the method of variation of parameters.

Solution

To solve the differential equation

y'' - y = e^x

using the method of variation of parameters: 1. **Solve the homogeneous equation

y'' - y = 0

:**

r^2 - 1 = 0 \implies r = \pm 1.

The general solution to the homogeneous equation is:

y_h = c_1 e^x + c_2 e^{-x}.

2. **Form the particular solution using variation of parameters:**

y_p = u_1(x) e^x + u_2(x) e^{-x}.

To find

u_1

and

u_2

, we solve:

u_1' e^x + u_2' e^{-x} = 0,

u_1' e^x - u_2' e^{-x} = e^x.

3. **Solve for

u_1'

and

u_2'

:**

u_1' e^x + u_2' e^{-x} = 0 \implies u_1' e^x = -u_2' e^{-x} \implies u_2' = -u_1' e^{2x}.

u_1' e^x - (-u_1' e^{2x}) = e^x \implies u_1' e^x + u_1' e^{2x} = e^x \implies u_1'(e^x + e^{2x}) = e^x \implies u_1' = \frac{e^x}{e^x + e^{2x}}.

u_1' = \frac{1}{1 + e^x} \implies u_1 = \int \frac{1}{1 + e^x} dx.

Using integration, we find:

u_1 = \ln|1 + e^x| + C.

4. **Substitute back to find

u_2

:**

u_2' = -u_1' e^{2x} = -\frac{e^{2x}}{1 + e^x}.

u_2 = \int -\frac{e^{2x}}{1 + e^x} dx.

5. **Solve the integrals and combine the solutions:**

y_p = u_1(x) e^x + u_2(x) e^{-x}.

6. **Result:** The general solution is:

y = y_h + y_p = c_1 e^x + c_2 e^{-x} + \text{(particular solution)}.

Solved on Solvelet with Basic AI Model

Some of the related questions asked by Alexander Brown on Solvelet

1. Use the variation of parameters to find a particular solution to the differential equation

y'' - y = 2x e^x

.2. Solve the initial value problem

y'' - y = 2x

y(0) = 1

y'(0) = 0

using the variation of parameters.,

DefinitionThe method of version of parameters is a way for finding particular solutions of non-homogeneous differential equations. This methods consists in the use of the solutions of the corresponding homogeneous equation to build the specific solution. as an example, in y′′+p(x)y′+q(x)y=g(x), the particular solution yp(x) represents what difference is needed from the complementary solution.

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